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let post_cover = article.cover--><div class="post_cover left_radius"><a href="/math/other/a_problem/" title="一个问题">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/math/other/a_problem/a_problem.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="一个问题"></a></div><div class="recent-post-info"><a class="article-title" href="/math/other/a_problem/" title="一个问题">一个问题</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-05-01T06:50:29.000Z" title="发表于 2021-05-01 14:50:29">2021-05-01</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/%E5%85%B6%E4%BB%96/">其他</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E5%AD%A6/">数学</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/math/other/a_problem/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">说在前面这个问题我还没有解决.
题目已知平面向量 $v_1,v_2,v_3,\cdots,v_n$ 满足 $\sum_{i=1}^n|\vec{v_i}|=1$ . 若必存在若干个向量, 它们的和的模不小于 $k$ , 求 $k$ 的最大值.
原题如下

给出的解法虽然很精彩, 但是并不是最大值 (最少是没证明 $1/4$ 是最大值) . 于是就产生了这个求最大值的想法.
猜想我猜测结果是 $1/\pi$ .
一些结果令 $P\subseteq {1,2,3,\cdots,n}$ , 且满足 $\left|\sum_{i\in P}v_i\right|$ 为所有向量相加的组合中能取到的最大值, 我们来考虑如何选取得到这个向量的集合. 设一个向量 $\vec{w}\not=\vec 0$ , 如何取向量并相加才能使得这个相加后的向量在 $\vec{w}$ 上的投影的模最大? 很显然, 取一条过原点的直线, 然后分别将直线的两边的所有向量相加 (在直线上的向量相加与否并不影响这个模的取值), 得到的两个向量必有一个在 $\vec{w}$ 上的投影的模是最大的, 但是这样取到的最大的模必然小 ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover right_radius"><a href="/math/other/an_inequality/" title="一个不等式">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/math/other/an_inequality/an_inequality.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="一个不等式"></a></div><div class="recent-post-info"><a class="article-title" href="/math/other/an_inequality/" title="一个不等式">一个不等式</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-04-25T04:16:46.000Z" title="发表于 2021-04-25 12:16:46">2021-04-25</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/%E5%85%B6%E4%BB%96/">其他</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E5%AD%A6/">数学</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E4%B8%8D%E7%AD%89%E5%BC%8F/">不等式</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/math/other/an_inequality/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content"> 发现一个不等式, 因此记录一下 (应该是正确的…吧)$$\sum_{i=1}^nk_i\sin(\theta_i)\geqslant\frac{\sqrt{2}}{2}\sin\left(\frac{\sum_{i=1}^nk_i\theta_i}{\sum_{i=1}^nk_i}\right)\sum_{i=1}^n k_i,\qquad \theta_i\in[0,\frac{\pi}{2}]$$对于 $\cos$ 函数也有类似的结果.
</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover left_radius"><a href="/math/other/complex_multiplication_and_rotation/" title="复数乘法与旋转">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/math/other/complex_multiplication_and_rotation/complex_multiplication_and_rotation.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="复数乘法与旋转"></a></div><div class="recent-post-info"><a class="article-title" href="/math/other/complex_multiplication_and_rotation/" title="复数乘法与旋转">复数乘法与旋转</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-04-17T13:11:45.000Z" title="发表于 2021-04-17 21:11:45">2021-04-17</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/%E5%85%B6%E4%BB%96/">其他</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E5%AD%A6/">数学</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E5%A4%8D%E6%95%B0/">复数</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/math/other/complex_multiplication_and_rotation/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">旋转公式对于一个点 $P_0(x_0,y_0)$ 来说 , $P_1(x_0\cos\theta-y_0\sin\theta,x_0\sin\theta+y_0\cos\theta)$ 是 $P_0$ 旋转 $\theta$ 角度 (顺时针) 后对应的点. 这个公式的推导是很简单的, 但是它这样的形式似乎并没有什么显而易见的规律. 但是, 复数乘法可以给我们一些启示.
复数乘法的几何意义设 $z_1=a_1+b_1\mathrm{i}, z_2=a_2+b_2\mathrm{i}$ , 那么就有 $z_1z_2=a_1a_2-b_1b_2+(a_1b_2+a_2b_1)\mathrm{i}$ . 如果还看不出什么, 就进行三角换元, 令 $a_1=k_1\cos\alpha, b_1=k_1\sin\alpha, a_2=k_2\cos\beta,b_2=k_2\sin\beta$ , 那么就有 $k_1=|z_1|, k_2=|z_2|$ , 于是$$\begin{aligned}z_1z_2&amp;=a_1a_2-b_1b_2+(a_1b_2+a_2b_1)\mathrm{i}\ ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover right_radius"><a href="/scribble/ggb_insert/" title="在网页中嵌入 GGB 最简单的方法">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/scribble/ggb_insert/ggb_insert.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="在网页中嵌入 GGB 最简单的方法"></a></div><div class="recent-post-info"><a class="article-title" href="/scribble/ggb_insert/" title="在网页中嵌入 GGB 最简单的方法">在网页中嵌入 GGB 最简单的方法</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-04-04T06:26:56.000Z" title="发表于 2021-04-04 14:26:56">2021-04-04</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%9D%82%E6%96%87/">杂文</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%99%E7%A8%8B/">教程</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/scribble/ggb_insert/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">一开始我是想要用官网的 js 解析 .ggb 文件的, 结果发现速度太慢还一堆问题… 最后发现直接分享然后 iframe 引用才是最快最方便的.
打开 GeoGebra , 然后在制作好的图中点击分享 (如下图)

然后就会进入登录界面, 有账号直接登录, 没有就注册.
接着进入 GGB 官网 , 登录并进入个人主页

接着看 资源 中是否有你分享的图, 如果没有, 重新回到 GGB 再次尝试分享.
然后点击进入你要嵌入到网页的资源, 查看该资源 ID .

记下 ID, 然后就可以在需要展示 GGB 的地方用 iframe 引用了
1234567&lt;iframe frameborder&#x3D;&quot;no&quot;title&#x3D;&quot;随便写&quot; src&#x3D;&quot;https:&#x2F;&#x2F;www.geogebra.org&#x2F;material&#x2F;iframe&#x2F;id&#x2F;刚刚记下的 ID&#x2F;smb&#x2F;false&#x2F;stb&#x2F;false&#x2F;stbh&#x2F;f ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover left_radius"><a href="/math/other/conical_section_conclusion/" title="圆锥曲线二级结论">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/math/other/conical_section_conclusion/conical_section_conclusion.gif" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="圆锥曲线二级结论"></a></div><div class="recent-post-info"><a class="article-title" href="/math/other/conical_section_conclusion/" title="圆锥曲线二级结论">圆锥曲线二级结论</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-03-21T03:32:14.000Z" title="发表于 2021-03-21 11:32:14">2021-03-21</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/%E5%85%B6%E4%BB%96/">其他</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E5%AD%A6/">数学</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E5%9C%86%E9%94%A5%E6%9B%B2%E7%BA%BF/">圆锥曲线</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/math/other/conical_section_conclusion/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">最近推了一个圆锥曲线的结论, 因此记录一下.
先上结论:
若有圆锥曲线 $\Gamma\colon\frac{x^2}{A}+\frac{y^2}{B}=1$ ,$A$ 与 $B$ 不同时取负数 (这样就同时代表了双曲线和椭圆) , 那么对于一个在该圆锥曲线上的点 $P_1(x_1,y_1)$ 来说, 存在一个点 $P_2(\frac{A-B}{A+B}x_1,\frac{B-A}{A+B}y_1)$ . 若作直线 $l_1, l_2$ 过点 $P_0$ 且相互垂直, 那么这两条直线与该圆锥曲线异于 $P_0$ 的两个交点 $M,N$ 的连线恒过点 $P_1$ .





对于形如 $y^2=2px$ 的抛物线也有类似结论, 不过 $P_2$ 为 $(2p+x_1,-y_1)$.




如何得到

我们以椭圆为例.

设椭圆 $\Gamma\colon\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , 点 $P_2 (x_2, y_2)$ , 直线 $l_{MN}\colon y-y_2=k(x-x_2)$  联立可得
$$
(a^2k^2+b^2)x^2+2 ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover right_radius"><a href="/hexo/hexo_conflict_mathjax/" title="解决 MathJax 与 Hexo 冲突问题">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/hexo/hexo_conflict_mathjax/hexo_conflict_mathjax.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="解决 MathJax 与 Hexo 冲突问题"></a></div><div class="recent-post-info"><a class="article-title" href="/hexo/hexo_conflict_mathjax/" title="解决 MathJax 与 Hexo 冲突问题">解决 MathJax 与 Hexo 冲突问题</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-02-20T07:08:10.000Z" title="发表于 2021-02-20 15:08:10">2021-02-20</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/Hexo/">Hexo</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/Hexo/">Hexo</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E6%B0%B4%E6%96%87/">水文</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/hexo/hexo_conflict_mathjax/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">将 Hexo 升级到 5.0 后, 发现 MathJax 与 Hexo 仍然有冲突. 然而网上大部分的方法对 5.0 版本的 Hexo 都没有用.
摸索许久后, 找到了一个比较不错的方法解决该问题. 即在公式前后加上 {% raw %} 与 {% endraw %}, 即
12345&#123;% raw %&#125;$$xxxxxx (公式内容)$$&#123;% endraw %&#125;

就可以完美解决.
这绝对不是水文.
</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover left_radius"><a href="/ML/RL/papers_read/Model-Free_RL/Deep_Q-Learning/1/" title="DQN 算法">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/ML/RL/papers_read/Model-Free_RL/Deep_Q-Learning/1/1.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="DQN 算法"></a></div><div class="recent-post-info"><a class="article-title" href="/ML/RL/papers_read/Model-Free_RL/Deep_Q-Learning/1/" title="DQN 算法">DQN 算法</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2021-02-13T14:55:54.000Z" title="发表于 2021-02-13 22:55:54">2021-02-13</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/">机器学习</a><i class="fas fa-angle-right"></i><a class="article-meta__categories" href="/categories/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/%E5%BC%BA%E5%8C%96%E5%AD%A6%E4%B9%A0/">强化学习</a><i class="fas fa-angle-right"></i><a class="article-meta__categories" href="/categories/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/%E5%BC%BA%E5%8C%96%E5%AD%A6%E4%B9%A0/%E5%85%A5%E9%97%A8/">入门</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/">机器学习</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E5%BC%BA%E5%8C%96%E5%AD%A6%E4%B9%A0/">强化学习</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E8%AE%BA%E6%96%87%E7%B2%BE%E8%AF%BB/">论文精读</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/ML/RL/papers_read/Model-Free_RL/Deep_Q-Learning/1/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">论文原文: Playing Atari with Deep Reinforcement Learning .
DQN (Deep Q-Networks) 算法, 简单来说就是 Deep Learning + Q Learning.
Q LearningQ Learning 实际上是维护一个 $Q$ 值表, 这个表可以认为是状态 $s$ 与动作 $a$ 的一个函数 $Q(s,a)$ , 其输出表示在状态 $s$ 下采用动作 $a$ 所获得的期望回报. 通过查询 $Q$ 值表, 就可以找出在某个状态 $s_t$ 下的最佳动作 $a_t$ .  Q Learning 算法的要点就在于得到一个足够真实的 $Q$ 值表. 而 $Q$ 值表的更新, 则基于 Bellman 方程 (详情可见RL 基本概念).
DQNDQN 的核心思想与 Q Learning 一样, 但区别就在于 $Q$ 函数. 在 Q Learning 中, $Q$ 函数是 $Q$ 值表, 也即一个个状态-动作对与期望回报的一一对应, 也就是离散的, 而在一些环境中 (比如 Atari 2600 games) 状态几乎是无穷多的, ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover right_radius"><a href="/scribble/player_in_blog/" title="为你的博客添加一个播放器">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/scribble/player_in_blog/player_in_blog.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="为你的博客添加一个播放器"></a></div><div class="recent-post-info"><a class="article-title" href="/scribble/player_in_blog/" title="为你的博客添加一个播放器">为你的博客添加一个播放器</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2020-11-29T08:05:16.000Z" title="发表于 2020-11-29 16:05:16">2020-11-29</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%9D%82%E6%96%87/">杂文</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%99%E7%A8%8B/">教程</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/Blog/">Blog</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/scribble/player_in_blog/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">话不多说, 直接开整.
播放器文件下载并引入地址在 https://github.com/cnyist/player .
下载下来后放入 source 目录中 (主题或 Hexo 自带的皆可) , 然后用 iframe 引入 index.html 文件
1iframe(name&#x3D;&quot;player&quot;, id&#x3D;&quot;player&quot;, src&#x3D;&quot;&#x2F;player&#x2F;&quot;, frameborder&#x3D;&quot;0&quot;, onload&#x3D;&quot;this.width&#x3D;player.jp_container_N.scrollWidth;$(&#39;.songlist__item&#39;).each(function()&#123;generate($(this).attr(&#39;id&#39;))&#125;);&quot;, style&#x3D;&quot;z-index:100; position:fixed; left:0px; bottom:0 ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover left_radius"><a href="/math/other/why_pi_not_4/" title="π 为什么会等于四">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/math/other/why_pi_not_4/1.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="π 为什么会等于四"></a></div><div class="recent-post-info"><a class="article-title" href="/math/other/why_pi_not_4/" title="π 为什么会等于四">π 为什么会等于四</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2020-10-18T08:07:34.000Z" title="发表于 2020-10-18 16:07:34">2020-10-18</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E5%AD%A6/">数学</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E6%9E%81%E9%99%90/">极限</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/math/other/why_pi_not_4/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">
我是在差不多三年前看到过这张图的. 曾经对极限完全不了解的我, 简单的认为由于边长一直不变, 所以不能这么算. 前几天有个同学的问题突然又激发了我对这个问题的思考, 突然意识到了之前的我对这个问题理解的浅显, 弄明白了这样论证的问题所在以及它为什么看起来如此正确.
为什么是错误的?首先, 我们自然知道 $\pi=3.1415926\cdots$ , 而上面的论证中不断的重复折叠所得到的周长, 最终都不会收敛于圆的周长 (一直等于 $4$) . 但是这里就会有人疑惑了: 为什么最终得到的 $4$ 不是圆的周长呢? 明明折到最后就是个圆呀? 而这, 其实源于我们对图中表述 不断的重复 和 一个图形怎样才是圆形 的理解的混乱所致.
什么是不断的重复?在数学中, 有一个对应的概念, 即为极限. 重复次数设为 $n$ , 然后 $\lim n\to\infty$ 就可以认为是不断的重复了.
一个图形怎样才是圆形?为了方便表述, 我们只考虑一个单位半圆 (半径为 $1$) 的圆弧

通过适当的建系, 可以通过一个函数来描述它, 我们用 $f(x)$ 来表示这个半圆. 现在, 我们列出两个关于判定 ...</div></div></div><div class="recent-post-item"><!-- - let post_cover = article.cover--><div class="post_cover right_radius"><a href="/ML/RL/papers_read/RL_papers_word/" title="强化学习类论文常用表达">     <img class="post_bg" src="https://cdn.jsdelivr.net/gh/cnyist/blog/ML/RL/papers_read/RL_papers_word/RL_papers_word.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="强化学习类论文常用表达"></a></div><div class="recent-post-info"><a class="article-title" href="/ML/RL/papers_read/RL_papers_word/" title="强化学习类论文常用表达">强化学习类论文常用表达</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time datetime="2020-09-13T06:13:47.000Z" title="发表于 2020-09-13 14:13:47">2020-09-13</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%9D%82%E6%96%87/">杂文</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/">机器学习</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E8%8B%B1%E8%AF%AD/">英语</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E5%BC%BA%E5%8C%96%E5%AD%A6%E4%B9%A0/">强化学习</a><span class="article-meta__link">•</span><a class="article-meta__tags" href="/tags/%E8%AE%BA%E6%96%87/">论文</a></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-comments"></i><a class="twikoo-count" href="/ML/RL/papers_read/RL_papers_word/#post-comment"></a><span class="article-meta-label"> 条评论</span></span></div><div class="content">由于自己英文实在是差, 所以把一些经常在论文里见到的却不太认得的表达记下来. 许多数学概念都有维基百科的相关链接, 要进去得 FQ , 如果你的浏览器可以装插件, 建议使用集装箱.



表达
词性
意义



probability distribution
n.
概率分布


discounted
adj.
有折扣的


finite
n.
有限的


infinite-horizon
n.
无限时间跨度


factor
n.
因子


denote
v.
表示


stochastic
adj.
随机的


standard definitions
n.
标准定义


unnormalized
adj.
非规范的


frequency
n.
频率


notation
n.
符号


imply
v.
暗示


nonnegative
adj.
非负的


guarantee
v.
保证


constant
adj.
不变的


deterministic
adj.
决定性


nonzero
adj.
非零的


algorithm
n.
算法


converge
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